4x4字母网格 - 字生成器 - 与朋友争夺

时间：2012-09-02 15:57:01

标签： ruby string algorithm grid permutation

我正在尝试基于4x4字母网格（下方）创建一个字生成器。

Scramble with Friends

以下是规则：

  • 不能重复的信件
  • 单词必须由相邻的字母组成
  • 单词可以水平，垂直或对角线形成左，右或上下

目前，我采用16个字符的输入并遍历字典中的每个单词，确定该单词是否可拼写为网格上的字母。 

#!/usr/bin/ruby

require './scores'   # alphabet and associated Scrabble scoring value (ie the wordValue() method)
require './words.rb' # dictionary of English words (ie the WORDS array)

# grab users letters
puts "Provide us the 16 letters of your grid (no spaces please)"
word = gets.chomp.downcase

arr = word.split('')

# store words that can be spelled with user's letters
success = []

# iterate through dictionary of words
WORDS.each do |w|

    # create temp arrays
    dict_arr = w.split('')
    user_arr = arr.dup
    test = true

    # test whether users letters spell current word in dict
    while test
        dict_arr.each do |letter|
            if (user_arr.include?(letter))
                i = user_arr.index(letter)
                user_arr.delete_at(i)
            else
                test = false
                break
            end
        end

        # store word in array
        if test 
            success << w
            test = false
        end
    end

end

# create hash for successful words and their corresponding values
SUCCESS = {}

success.each do |w|
  score = wordValue(w)
  SUCCESS[w] = score
end

# sort hash from lowest to smallest value
SUCCESS = SUCCESS.sort_by {|word, value| value}

# print results to screen
SUCCESS.each {|k,v| puts "#{k}:  #{v}"}

但是，这种方法没有考虑到电路板上瓷砖的位置。 您如何建议我根据他们在4x4网格中的位置找到可以创建的单词？

对于上图中的棋盘游戏，我的VM运行Ubuntu大约需要1.21秒来计算1185个可能的单词。我在/ usr / share / dict / words

中使用Ubunut提供的词典

4 个答案:

答案 0 :(得分：3)

不是迭代单词并搜索它们的存在，而是遍历网格上的每个图块并找到源自该图块的所有单词。

首先，将字典编译成trie。尝试在执行前缀匹配字符串比较时非常有效，这对我们很快就会有用。

要查找电路板中的字，请对16个图块中的每一个执行以下步骤，从prefix的空字符串开始。

  1. 将当前图块的值添加到prefix。
  2. 检查我们的trie是否包含以prefix开头的任何字词。
  3. 如果是，请对搜索进行分支：对于与此图块相邻的每个合法（未访问）图块，请返回步骤1（递归）。
  4. 如果不匹配，请停止此搜索分支，因为没有匹配的单词。

答案 1 :(得分：1)

我会创建一个代表整个电路板的简单图表。字母是顶点。如果两块字母在板上彼此靠近，我会在它们的顶点之间创建一条边。很容易找出输入是否有效。您只需检查图表中是否存在匹配的路径。

答案 2 :(得分：0)

我原来的答案不是你想要的。我正在创建所有的列表 网格中的“单词”，而不是搜索您已识别的单词 来自字典。现在我编写了一个搜索网格的函数 特别的话。它以递归方式工作。

所以，现在算法是：

1）获取用户的16个字母
2）搜索具有这些字母的所有单词的字典
3）使用每个单词调用is_word_on_board以查看是否有匹配

#!/usr/bin/ruby

# This script searches a board for a word
#
# A board is represented by a string of letters, for instance, the string
# "abcdefghijklmnop" represents the board:
#
#    a b c d
#    e f g h
#    i j k l
#    m n o p
#
# The array ADJACENT lists the cell numbers that are adjacent to another
# cell.  For instance ADJACENT[3] is [2, 6, 7].  If the cells are numbered
#
#     0  1  2  3
#     4  5  6  7
#     8  9 10 11
#    12 13 14 15

ADJACENT = [
    [1, 4, 5],
    [0, 2, 4, 5, 6],
    [1, 3, 5, 6, 7],
    [2, 6, 7],
    [0, 1, 5, 8, 9],
    [0, 1, 2, 4, 6, 8, 9, 10],
    [1, 2, 3, 5, 7, 9, 10, 11],
    [2, 3, 6, 10, 11],
    [4, 5, 9, 12, 13],
    [4, 5, 6, 8, 10, 12, 13, 14],
    [5, 6, 7, 9, 11, 13, 14, 15],
    [6, 7, 10, 14, 15],
    [8, 9, 13],
    [8, 9, 10, 12, 14],
    [9, 10, 11, 13, 15],
    [10, 11, 14]
]

# function:  is_word_on_board
#
# parameters:
#   word   - word you're searching for
#   board  - string of letters representing the board, left to right, top to bottom
#   prefix - partial word found so far
#   cell   - position of last letter chosen on the board
#
# returns true if word was found, false otherwise
#
# Note:  You only need to provide the word and the board.  The other two parameters
# have default values, and are used by the recursive calls.

# set this to true to log the recursive calls
DEBUG = false

def is_word_on_board(word, board, prefix = "", cell = -1)
    if DEBUG
        puts "word = #{word}" 
        puts "board = #{board}"
        puts "prefix = #{prefix}"
        puts "cell = #{cell}"
        puts
    end

    # If we're just beginning, start word at any cell containing
    # the starting letter of the word
    if prefix.length == 0
        0.upto(15) do |i|
            if word[0] == board[i]
                board_copy = board.dup
                newprefix = board[i,1]

                # put "*" in place of letter so we don't reuse it
                board_copy[i] = ?*

                # recurse, and return true if the word is found
                if is_word_on_board(word, board_copy, newprefix, i)
                    return true
                end
            end
        end

        # we got here without finding a match, so return false
        return false
    elsif prefix.length == word.length
        # we have the whole word!
        return true
    else
        # search adjacent cells for the next letter in the word
        ADJACENT[cell].each do |c|
            # if the letter in this adjacent cell matches the next
            # letter of the word, add it to the prefix and recurse
            if board[c] == word[prefix.length]
                newprefix = prefix + board[c, 1]
                board_copy = board.dup

                # put "*" in place of letter so we don't reuse it
                board_copy[c] = ?*

                # recurse, and return true if the word is found
                if is_word_on_board(word, board_copy, newprefix, c)
                    return true
                end
            end
        end

        # bummer, no word found, so return false
        return false
    end
end

puts "Test board:"
puts
puts "  r u t t"
puts "  y b s i"
puts "  e a r o"
puts "  g h o l"
puts

board = "ruttybsiearoghol"

for word in ["ruby", "bears", "honey", "beast", "rusty", "burb", "bras", "ruttisbyearolohg", "i"]
    if is_word_on_board(word, board)
        puts word + " is on the board"
    else
        puts word + " is NOT on the board"
    end
end

运行此脚本会产生以下结果：

Test board:

  r u t t
  y b s i
  e a r o
  g h o l

ruby is on the board
bears is on the board
honey is NOT on the board
beast is on the board
rusty is NOT on the board
burb is NOT on the board
bras is on the board
ruttisbyearolohg is on the board
i is on the board

答案 3 :(得分：0)

我碰巧有一个我刚才写过的Boggle求解器。它遵循Cheeken的大纲。它的调用方式略有不同（你提供单词列表文件和一个带有4x4网格作为参数的文本文件），但我认为值得分享。另请注意，它会对待＆＃34; Q＆＃34;作为＆＃34; QU＆＃34;，所以那里有一些额外的逻辑。

require 'set'

def build_dict(dict, key, value)
  if key.length == 0
    dict[:a] = value
  else
    if key[0] == "q"
      first = key[0..1]
      rest = key[2, key.length - 1]
    else
      first = key[0]
      rest = key[1, key.length - 1]
    end

    dict[first] = {} unless dict.has_key? first
    build_dict(dict[first], rest, value)
  end
end

dict = {}
#parse the file into a dictionary
File.open(ARGV[0]).each_line do |line|
  real_line = line.strip
  build_dict(dict, real_line, real_line)
end

#parse the board
board = {}
j = 0
File.open(ARGV[1]).each_line do |line|
  line.chars.each_with_index do |l, i|
    board[[j, i]] = l
  end
  j += 1
end

#(0..3).each{|r| puts (0..3).map{|c| board[[r, c]]}.join}

#how can i get from one place to another?
def get_neighbors(slot, sofar)
  r, c = slot
  directions =
   [
    [r+1, c],
    [r+1, c+1],
    [r+1, c-1],
    [r, c+1],
    [r, c-1],
    [r-1, c],
    [r-1, c+1],
    [r-1, c-1]
   ]
  directions.select{|a| a.all?{|d| d >= 0 && d <= 3} && !sofar.include?(a)}
end

#actual work
def solve(board, slot, word_dict, sofar)
  results = Set.new
  letter = board[slot]
  letter = "qu" if letter == "q"
  stuff = word_dict[letter]
  return results if stuff.nil?
  if stuff.has_key? :a
    results << stuff[:a] if stuff[:a].length > 2
  end
  unless stuff.keys.select{|key| key != :a}.empty?
    get_neighbors(slot, sofar).each do |dir|
      results += solve(board, dir, stuff, sofar.clone << slot)
    end
  end
  results
end

#do it!
results = Set.new
all_slots = (0..3).to_a.product((0..3).to_a)
all_slots.each do |slot|
  results += solve(board, slot, dict, slot)
end

puts results.sort
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